As a cold air mass advances on a warm air mass, what usually comes before it?
1 answer:
You might be interested in
Answer:
b. 1.1 m
Explanation:
It is given that the total distance between the masses is equal to the length of the board, which is 3 m. Therefore,
where,
s₁ = distance of fulcrum from left mass
s₂ = distance of fulcrum from right mass
In order to achieve balance, the torque due to both masses must be equal:
s₁ = 1.1 m
Hence, the correct option is:
<u>b. 1.1 m</u>
Answer:
θ = 12.95º
Explanation:
For this exercise it is best to separate the process into two parts, one where they collide and another where the system moves altar the maximum height
Let's start by finding the speed of the bar plus clay ball system, using amount of momentum
The mass of the bar (M = 0.080 kg) and the mass of the clay ball (m = 0.015 kg) with speed (v₀ = 2.0 m / s)
Initial before the crash
p₀ = m v₀
Final after the crash before starting the movement
= (m + M) v
p₀ =
m v₀ = (m + M) v
v = v₀ m / (m + M)
v = 2.0 0.015 / (0.015 +0.080)
v = 0.316 m / s
With this speed the clay plus bar system comes out, let's use the concept of conservation of mechanical energy
Lower
Em₀ = K = ½ (m + M) v²
Higher
= U = (m + M) g y
Em₀ =
½ (m + M) v² = (m + M) g y
y = ½ v² / g
y = ½ 0.316² / 9.8
y = 0.00509 m
Let's look for the angle the height from the pivot point is
L = 0.40 / 2 = 0.20 cm
The distance that went up is
y = L - L cos θ
cos θ = (L-y) / L
θ = cos⁻¹ (L-y) / L
θ = cos⁻¹-1 ((0.20 - 0.00509) /0.20)
θ = 12.95º